以相反的顺序输出数组的元素

Question

User puts unknown amount of integers in an array (dynamic). Program has to output those numbers in reverse order, for instance: Integers in reverse order. Exmp. 324,12,5987 --> 423, 21, 7895 Exmp. 123, 100 --> 321, 1 (two 0's get removed)

I'm not sure if I'm on the right path, but I sincerely hope that someone could suggest me a solution. Right now, I get the error of ''Segmentation fault'' - could it be related with array space?

Here's my code for the moment:

int main()
{

    int n,sk,rez;


    int *a;


    cout << "How many integers will there be? : " << endl;
    cin >> sk;

    a = new int[sk];


    for (int i = 0; i < sk; i++)
    {

        cout << "Write an integer: " << endl;
        cin >> n;
        a[i] = n;
    }


    for (int i = 0; i < sk; i++)
    {
        // Gets amount of the digits of every integer.

        int count, new_num = 0;
        int* skaits;

        while(a[i] > 0)
        {
        count++;
        a[i] = a[i]/10;
        }
        skaits = new int [count];
        for (int j = 0; j<count; j++)
            {
                skaits[j] = count; //  Exmp. I write 324,12 , loop calculates 3 and 2 and puts that amount in an array.

                // Loop to output integers in reverse order. Exmp. 324,12,5987 --> 423, 21, 7895

                for (int k = 0; k<skaits[j]; k++)
                {
                    new_num = new_num * 10 + (a[k] % 10);
                    a[k] = a[k]/10;


                }
                cout << new_num << endl;
            }

    }



    delete []a;
    return 0;

}

Answer 1

As a general rule: try to make your code more modular. I would expect to find a function which computes the reverse of a single number. This is easier to write and much more easier to debug, since you separate the two tasks: collect data, reverse the integer.

Anyway, looking deeper in your code: it seems that after you have computed count , the content of the variable a[i] has been destroyed... ( a[i]==0 to exit the while loop).

A simple way to reverse a decimal number:

int reverse(int n) {
    int m=0;
    while (n>0) {
        m *= 10;
        m += n%10;
        n /= 10;
    }
    return m;
}

Answer 2

Here segmentation fault occurs as count is not initialized. I have found many errors in your program, for instance:

 int count, new_num = 0;// here count is not initialized also need count=0;
        int* skaits;

        while(a[i] > 0)
        {
        count++; // without initialize how you increase count 
        a[i] = a[i]/10;
        }// after this your number in a[i] is become 0. so what will you find

      skaits = new int [count];// no need extra skaits. you have count already

It is better you can use a function to reverse each number of a[i] .

long reverse(long n) {
   static long r = 0;

   if (n == 0) 
      return 0;

   r = r * 10;
   r = r + n % 10;
   reverse(n/10);
   return r;
} 

And you can do like as follows:

for(int i=0;i<sk;i++)
 {
  int new_num=reverse(a[i);
   a[i]=new_num;
   cout<<a[i]<<endl;
}

Or you can just do it just like as follows:

for (int i = 0; i < sk; i++)
    {

        cout << "Write an integer: " << endl;
        cin >> n;
        a[i] = n;
    }
    for (int i = 0; i < sk; i++)
    {
        // Gets amount of the digits of every integer.
        int reverse=0;
        while (a[i] != 0)
        {
            reverse = reverse * 10;
            reverse = reverse + a[i]%10;
            a[i]= a[i]/10;
        }
        cout<<reverse<<endl;

    }

An another way to do this, first make each number int to string first and then reverse it.